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-rw-r--r--prolog/problems/sorting/is_sorted_1/en.py81
-rw-r--r--prolog/problems/sorting/is_sorted_1/sl.py2
2 files changed, 81 insertions, 2 deletions
diff --git a/prolog/problems/sorting/is_sorted_1/en.py b/prolog/problems/sorting/is_sorted_1/en.py
index 8741bc7..678494c 100644
--- a/prolog/problems/sorting/is_sorted_1/en.py
+++ b/prolog/problems/sorting/is_sorted_1/en.py
@@ -10,4 +10,83 @@ description = '''\
false.
</pre>'''
-hint = {}
+plan = ['''\
+<p>As always try to reduce the problem onto a smaller one. Do an appropriate check (comparison) at
+the start of the list, and submit the tail to a recursive check.</p>
+''', '''\
+<p>You do know how to access the first <em>two</em> elements of the list, right? And the arithmetic comparison
+operators were introduced in the previous batch of exercises.</p>
+''', '''\
+<p>If the given list <code>L</code> is composed of heads <code>H1</code> and <code>H2</code> and of tail
+<code>T</code>, and if we assume that tail <code>T</code> along with the second head is sorted, and furthermore
+<code>H1</code> is smaller or equal to <code>H2</code>, then the whole list <code>L</code> is sorted.</p>
+''']
+
+hint = {
+ 'eq_instead_of_equ': '''\
+<p>The operator <code>==</code> is "stricter" than operator <code>=</code> in the sense that
+for the latter it is enough to be able to make the two operands equal (unification).</p>
+<p>Of course, you can also solve the exercise without explicit use of either of these two operators, just
+remember that unification is implicitly performed with the predicate's arguments (head of clause).</p>
+''',
+
+ 'eq_instead_of_equ_markup': '''\
+<p>Perhaps the operator for unification (=) would be better?</p>
+''',
+
+ 'base_case': '''\
+<p>Did you think of a base case? Which is one of the shortest sorted lists?</p>
+''',
+
+ 'recursive_case': '''\
+<p>The base case is ok. However, what about the general recursive case?</p>
+''',
+
+ 'predicate_always_false': '''\
+<p>It seems your predicate is <em>always</em> "false". Did you give it the correct name,
+or is it perhaps misspelled?</p>
+<p>If the name is correct, check whether something else is misspelled, perhaps there is a full stop instead of
+a comma or vice versa, or maybe you typed a variable name in lowercase?</p>
+<p>It is, of course, also possible that your conditions are too restrictive, or even impossible to satisfy
+(as would be, for example, the condition that <code>X</code> is <em>simultaneously</em> smaller and greater than
+<code>Y</code>, or something similarly impossible).</p>
+''',
+
+ 'timeout': '''\
+<p>Is there an infinite recursion at work here? How will it ever stop?</p>
+<p>Or perhaps is there a missing, faulty, or simply incompatible (with the general recursive case) base case?</p>
+''',
+
+ '[]_base_case_missing': '''\
+<p>For completeness, please take care of the special case, i.e. the empty list which is also sorted.
+But be careful not to destroy the other solutions.</p>
+''',
+
+ 'duplicates_fail': '''\
+<p>Did you forget about duplicate elements? The list below is also sorted!</p>
+<p><code>?- is_sorted([25,25,25,25]).</code></p>
+''',
+
+ 'H1_instead_of_H2_sent_into_recursion': '''\
+<p>Did you send the wrong of the two list heads into recursion?</p>
+''',
+
+ 'base_case_at_len_1_missing': '''\
+<p>The recursive case in this exercise requires two elements, even though you put one back when the tail
+is recursively processed. But what happens when there is just a single element left in the list? Think
+about it as this will probably be the main base case.</p>
+''',
+
+ 'both_heads_omitted_from_recursion': '''\
+<p>Are you taking two elements out of the list before initiating a recursive call? You're trying to
+reduce the number of comparisons, aren't you? ;) But unfortunately this will not be possible! Of the following
+two cases below one works correctly and one doesn't. What's the difference between them?</p>
+<p><code>?- is_sorted([1,3,14,16,24,25]).</code></p>
+<p><code>?- is_sorted([24,25,14,16,1,3]).</code></p>
+''',
+
+ 'min_used': '''\
+<p>Try solving this exercise without using the predicate <code>min/2</code>. Your solution should have the
+time complexity of O(n). By using <code>min/2</code> the complexity is typically about O(n*n).</p>
+''',
+}
diff --git a/prolog/problems/sorting/is_sorted_1/sl.py b/prolog/problems/sorting/is_sorted_1/sl.py
index de52984..36ace61 100644
--- a/prolog/problems/sorting/is_sorted_1/sl.py
+++ b/prolog/problems/sorting/is_sorted_1/sl.py
@@ -77,7 +77,7 @@ glavni robni pogoj!</p>
'both_heads_omitted_from_recursion': '''\
<p>Jemlješ po dva elementa iz seznama preden greš v rekurzijo? Poskušaš malce prihraniti pri primerjavah, kajne? ;)
-Ampak žal to ne gre! Od spodnjih dveh primerov eden deluje pravilno in eden ne, ugotovi v čem je razlika. </p>
+Ampak žal to ne gre! Od spodnjih dveh primerov eden deluje pravilno in eden ne, ugotovi v čem je razlika.</p>
<p><code>?- is_sorted([1,3,14,16,24,25]).</code></p>
<p><code>?- is_sorted([24,25,14,16,1,3]).</code></p>
''',